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spayn [35]
3 years ago
15

As an object falls freely toward the earth, the momentum of the object-earth system (1) decreases (2) increases (3) remains the

same ...?
Physics
1 answer:
Brums [2.3K]3 years ago
4 0

The best and most correct answer among the choices provided by your question is the third choice or number 3.

<span>As an object falls freely toward the earth, the momentum of the object-earth system remains the same.</span>


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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An airplane travels at 300 mi/h south for 2.00 h and then at 250 mi/h north for 750 miles. What is the average speed for the tri
Pie
Avg sped = total distance/ total time = 1350 mi/ 5 hrs= 270mph (i dont know if ur teacher wants you to convert this to m/s) 

300miles are traveled in 1 hr. So, 300 *2hrs = 600 miles south
750/250= 3hrs north
Total distance = 600 miles + 750 miles= 1350 miles
Total time is =  3hrs + 2hrs= 5hrs
 



7 0
3 years ago
A fairgrounds ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follo
cestrela7 [59]

Answer:

13.37 rev/min

Explanation:

acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration (a_c) = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².

r = 9 m

Centripetal acceleration (a_c) is given by:

a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s

The velocity (v) is given by:

v = ωr;  where ω is the angular velocity

Hence:

ω = v/r = 12.6 / 9

ω = 1.4 rad/s

ω = 2πN

N = ω/2π = 1.4 / 2π

N = 0.2228 rev/s

N = 13.37 rev/min

4 0
2 years ago
The work function for magnesium is 3.70 ev. what is its cutoff frequency?
alexandr402 [8]

The cutoff frequency for magnesium is 8.93 x 10¹⁴ Hz.

<h3>What is cutoff frequency?</h3>

The work function is related to the frequency as

W0 = h x fo

where, fo = cutoff frequency and h is the Planck's constant

Given is the work function for magnesium is  3.70 eV.

fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴

fo = 8.93 x 10¹⁴ Hz.

Thus, the cut off frequency is 8.93 x 10¹⁴ Hz.

Learn more about cutoff frequency.

brainly.com/question/14378802

#SPJ1

7 0
2 years ago
What is the formula for Ba+2 and F-?
Alinara [238K]

Answer:

BaF2

Explanation:

since you got the valence numbers just do the scissors move where you:

give the F the 2 and give the Ba the 1 so it be like

BaF2 here is the chemical compound

5 0
2 years ago
If gravity on the earth increased, what affect would it have on the moon
Rufina [12.5K]

Answer:

If gravity on Earth is increased, this gravitational tugging would have influenced the moon's rotation rate. If it was spinning more than once per orbit, Earth would pull at a slight angle against the moon's direction of rotation, slowing its spin. If the moon was spinning less than once per orbit, Earth would have pulled the other way, speeding its rotation.

6 0
3 years ago
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